PDF chapter test TRY NOW
The following steps can determine the median of the grouped frequency distribution:
1. Find the cumulative frequency distribution and denote \(n\) as the total frequency.
2. Find the \(\frac{n}{2}^{th}\) term.
3. The class which contains the cumulative frequency \(\frac{n}{2}\) is the median class.
4. The median of the class can be determined using the formula:
Median \(= l + \frac{(\frac{n}{2} - cf)}{f} \times h\)
Where \(l\) is the lower limit of the median class,
\(cf\) is the cumulative frequency of the class preceding the median class,
\(f\) is the frequency of the median class,
\(h\) is the width of the median class, and
\(n\) is the total frequency.
Example:
Find the median of the total marks(out of \(500\)) scored by students in a school of class \(X\). Also, construct the cumulative frequency distribution of less than type and cumulative frequency distribution of more than type.
| Total marks | \(100 - 200\) | \(200 - 300\) | \(300 - 400\) | \(400 - 500\) |
| No. of students | \(30\) | \(15\) | \(26\) | \(44\) |
Solution:
| Total marks | No. of students | Cumulative frequency |
| \(100 - 200\) | \(30\) | \(30\) |
| \(200 - 300\) | \(15\) | \(45\) |
| \(300 - 400\) | \(26\) | \(71\) |
| \(400 - 500\) | \(44\) | \(115\) |
Therefore, the total frequency is \(n = 115\).
Median class \(= (\frac{n}{2}^{th})\) value
\(= (\frac{115}{2})^{th}\) value
\(= (57.5)^{th}\) value
The median of the grouped frequency distribution can be determined using the formula \(l + \frac{(\frac{n}{2} - cf)}{f} \times h\)
The value \(57.5\) lies in the class interval \(300 - 400\)
Here, \(l = 300\), \(\frac{n}{2} = 57.5\), \(cf = 45\), \(f = 26\) and \(h = 100\)
Substituting the known values in the above formula, we have:
Median \(= 300 + (\frac{57.5 - 45}{26}) \times 100\)
\(= 300 + (0.481) \times 100\)
\(= 300 + 48.1\)
\(= 348.1\)
Therefore, the median of the given data is \(348.1\)
From the above data, we can see that the number of students who scored less than \(200\) marks is \(30\).
Can you able to find the number of students who scored less than \(400\) marks?
The answer is yes. We can find the number of students by adding the frequencies of the class intervals \(100 - 200\), \(200 - 300\) and \(300 - 400\). Thus, the number of students are \(30 + 15 + 26 = 71\).
Similarly, we can determine the cumulative frequencies of other classes.
| Total marks | Cumulative frequency |
| Less than \(200\) | \(30\) |
| Less than \(300\) | \(30 + 15 = 45\) |
| Less than \(400\) | \(45 + 26 = 71\) |
| Less than \(500\) | \(71 + 44 = 115\) |
This type of distribution is called cumulative frequency distribution of less than type.
Similarly, let us construct the cumulative frequency distribution of more than type for the given table.
From the given question, we observe that the total number of students is \(115\). Let us start with the cumulative frequency of \(115\) since the number of students who scored more than or equal to \(100\) is \(115\).
| Total marks | Cumulative frequency |
| More than or equal to \(100\) | \(115\) |
| More than or equal to \(200\) | \(115 - 30 = 85\) |
| More than or equal to \(300\) | \(85 - 15 = 70\) |
| More than or equal to \(400\) | \(70 - 26 = 44\) |
Here, \(100\), \(200\), \(300\), and \(400\) denote the lower limits of the respective class intervals.
Register for free to see more content